Serbian Mathematical Olympiad

Solution:

Let us denote by $K$ the point of tangency of line $t$ and circle $BOC$. Let the circles circumscribed around triangles $BDK$ and $CEK$ intersect at point $X \neq K$. Since $\text{B X C = B X K + K X C = A D K + K E A = 180 - C A B}$, point $X$ lies on the circumscribed circle $k$ of triangle $A B C$. Moreover, $\text{D X E = D X K + K X E = D B K + K C E = C K B - C A B = C A B = D X E}$, so $X$ also lies on the circumscribed circle $k_1$ of triangle $A' D E$. We will prove that circles $k$ and $k_1$ are tangent at point $X$.

If lines $C K$ and $X D$ intersect at point $P$, then $\text{X P C = X D E - C K E = X B K - C B K = X B C}$, which means that $P$ lies on circle $k$. Analogously, lines $B K$ and $X E$ intersect at point $Q$ on circle $k$. Finally, from $\text{X P Q = X B Q = X D K}$ it follows that

that $P Q \parallel D E$. Therefore, triangles $X D E$ and $X P Q$ are homothetic with center of homothety $X$, so their circumscribed circles are tangent at $X$.

Solution:

Second solution. Let lines $B K$ and $C K$ intersect the circumscribed circle of $\triangle A B C$ again at points $Q$ and $P$, respectively. From $\text{C P Q = C B Q = C K E}$ it follows that $P Q \parallel D E$. Let lines $D P$ and $E Q$ intersect at point $X$. Since points $D = P X \cap A B$, $K = P C \cap Q B$ and $E = A C \cap Q X$ are collinear, by the converse of Pascal's theorem point $X$ lies on the same circle as points $A, B, C, P, Q$. Therefore, triangles $X D E$ and $X P Q$ are homothetic, so their circumscribed circles are tangent at their center of homothety $X$. Finally, point $A'$ lies on the circumscribed circle of $\triangle D E X$ because $\text{D X E = P X Q = P C A + A B Q = B K C - B A C = B A C = D A' E}$.